Schoolcrop

(Q5) In △ACB, ∠C = 90° and CD ⊥ AB. Prove that
BC²

AC²
=
BD

AD

Solution :

Proof : △ADC and △CDB are similar.

∴

∆

∆

=

( ) × AD × DC

( ) × BD × DC

=

AD

BD

⇒

∆

∆

=

BD

AD

.....(1)

or

∆

∆

=

BC²

AC²

.....(2)

[Ratio of areas of similar triangles is equal to the ratio of squares of their corresponding sides.]

From (1) and (2),

BD

AD

=

BC²

AC²